# Hi Jadarius,

The height h of a projectile above the ground after t seconds is given by :

** h(t) = - (1 / 2) g t**^{ 2}** + v**_{o}** t + h**_{o}

g is the acceleration due to gravity which on earth is approximately equal to 32 feet / s^{2}

then **h(t) = - (1 / 2)(**32**) t**^{ 2}** + v**_{o}** t + h**_{o}

** h(t) = ****- 16**** t**^{ 2}** + v**_{o}** t + h**_{o}

**Since the rock was thrown from an initial height of 192 feet then ****h**_{o}**=192**

**the rock reached its maximum height of 256 feet after 2 seconds**

then ** h(t) = ****- 16**** t**^{ 2}** + v**_{o}** t + 192 **

For a quadratic function of the form h = a t^{ 2} + b t + c, the vertex is located at t = - b / 2a

and since a is negative, the maximum value of h occur at t = - b / 2a (vertex of the graph of h(t)).

**since the rock reached its maximum height of 256 feet after 2 seconds**

then 2 = - b / 2(-16)

2 = - b / -32

so b = 64

which means our initial velocity **v**_{o}** = 64**

then our equation becomes** ****h(t) = ****- 16**** t**^{ 2}** + 64 t + 192 **

** to check if the equation is correct** we can simply plug in

**t = 2**and see if it will give us the

**maximum height of 256:**

**h(t) = ****- 16**** (2)**^{ 2}** + 64 (2) + 192 = = 256 **

**so now we can find after ****how many seconds(t=?) the rock hit the ground (height = 0)**

** - 16**** t**^{ 2}** + 64 t + 192 = 0 **

so to solve we need to factor out -16 as a GCF and divide the whole equation by -16 which will result in:

** t**^{ 2}** - 4 t - 12 = 0 **

now we can factor the trinomial on the left side of the equation:

**( t - 6 ) ( t + 2 ) = 0**

t - 6 = 0 or t + 2 = 0

+6 +6 - 2 -2

** t = 6** t = -2 ( not possible since time cannot be negative)

Now to ** check our answer** we can simply plug in

**t = 6**in the equation

**- 16**

**t**

^{ 2}

**+ 64 t + 192 = 0**

**-16 (6)**^{2}** + 64(6) + 192 = 0**

** 0 = 0**

so the rock will hit the ground after** 6 Seconds**

I hope this helps, please feel free to ask any questions regarding this solution!

Best Regards.